$y= 0.4\ x^2 \cdots (Given)\ Hence,\ \dfrac {dy} {dx}= 0.8x\ and \dfrac {d^2y} {dx^2}=0.8\ At\ x=2, \dfrac {dy} {dx}= 0.8 \times 2= 1.6\ and \dfrac {d^2y} {dx^2}=0.8\$

$Radius\ of\ curvature\ (\rho)= {\bigg [1+ \ (dy/dx)^2 \bigg] ^{3/2} \over{d^2y / dx^2}}\\ {\bigg [1+ (1.6)^2 \bigg] ^{3/2} \over{0.8}}\\ = 8.3962 \ m. $

$x=2,\ velocity (v)= 6 m/s\ and\ tangential\ acceleration\ (a_t)=3 m/s^2\cdots (Given)\\ With\ the\ horizontal,\ let\ velocity\ vector\ make\ an\ angle\ \theta\\ Hence,\ tan\theta =m=slope\ of\ tangent= \dfrac {dy} {dx}= 1.6 $

$\theta=58\ ^\circ\\ Velocity\ along\ x-direction\ (v_x)= vcos\theta=6.cos58 = 3.1795 \ m/s\\ Velocity\ along\ y-direction\ (v_y)= vsin\theta=6.sin58 = 5.0883 \ m/s\\ $

$Normal\ acceleration\ (a_n)=\dfrac {v^2} {\rho}= \dfrac {6^2} {8.3962}\\ =4.2877\ m/s^2\\ a^2= {a_1}^2+{a_2}^2 = 3^2 + 4.2877^2 = 27.3844\\ Hence,\ (a)=5.2330\ m/s^2.\\ $

$Let\ acceleration\ vector\ make\ an\ angle\ '\alpha'\ with\ a_t.\\ Hence,\ \alpha=tan^{-1} \dfrac {a_n} {a_t}\\ =tan^{-1} \dfrac {4.2877} 3\\ = 55.0204 \ ^\circ\\ With\ horizontal,\ angle\ made\ by\ acceleration\ = \theta + \alpha\\ =55.0204 + 58 \\ = 113.0204 ^\circ$

$Hence, Velocity\ along\ x-direction\ (v_x)= 3.1795 \ m/s;\ Velocity\ along\ y-direction;\ (v_y)= 5.0883 \ m/s\ and \ acceleration=5.2330\ m/s^2, \angle=113.0204\ ^\circ$